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So we get
∠ BAE = ∠ CAF = ∠ A
Therefore, by ASA congruence criterion we get
△ ABE ≅△ ACF
(ii) As we know that △ ABE ≅△ ACF
It is proved that AB = AC.

Question 4
△ABC and △DBC are two isosceles triangles on the same base BC and vertices A
and D are on the same side of BC. If AD is extended to interest BC at E, show that
(i) △ABD ≅△ACD
(ii) △ABE ≅△ACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.

(i) Based on the △ ABD and △ ACD
From △ ABC we know that AB and AC are equal sides of isosceles triangle
So we getAB = AC
From △ DBC we know that DB and DC are equal sides of isosceles triangle
So we getDB = DC
We also know that AD is common i.e. AD = AD
Therefore, by SSS congruence criterion we get
△ ABD ≅△ ACD

(ii) We know that △ ABD ≅△ ACD
We get ∠ BAD = ∠ CAD (c. p. c. t)
It can be written as
∠ BAE = ∠ CAE ……… (1)
Considering △ ABE and △ ACE
We know that AB and AC are the equal sides of isosceles △ ABC
AB = AC
So by using equation (1) we get∠ BAE = ∠ CAE
We know that AE is common i.e. AE = AE
Therefore, by SAS congruence criterion we get
△ ABE ≅△ ACE

(iii) We know that △ ABD ≅△ ACD
We get ∠ BAD = ∠ CAD (c. p. c. t)
It can be written as∠ BAE = ∠ CAE
Therefore, it is proved that AE bisects ∠ A.